Difference between revisions of "2023 AMC 10A Problems/Problem 6"
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~Mintylemon66 | ~Mintylemon66 | ||
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| + | ==Solution 2== | ||
| + | Just set one vertice equal to <math>21</math>, it is trivial to see that there are <math>3</math> faces with value <math>42</math>, and <math>42 \cdot 3=126</math>. | ||
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| + | ~SirAppel | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2023|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:48, 9 November 2023
Contents
Problem
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is 
. What is the value of the cube?
Solution
Each of the vertices is counted 
times because each vertex is shared by three different edges.
Each of the edges is counted 
times because each edge is shared by two different faces.
Since the sum of the integers assigned to all vertices is , the final answer is 
~Mintylemon66
Solution 2
Just set one vertice equal to , it is trivial to see that there are faces with value 
, and 
.
~SirAppel
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
